Photon entanglement: why three angles?

[RUTA]

So those are counts of matching and mismatching pairs. Equation 20 is then the same what I called correlation formula: match - mismatch, and QM's cos^2(theta). It means two uniformly random binary sequences should be completely uncorrelated with equal number of matches and mismatches.

0.5:0.5 vs 0.5:0.5
chance for match: (0.5 * 0.5) + (0.5 * 0.5) = 0.5
chance for mismatch: (0.5 * 0.5) + (0.5 * 0.5) = 0.5
correlation: 0.5 - 0.5 = 0%

I don't know what you're talking about, sorry. You have completely correlated outcomes for α-β = 0 (P_VV = P_HH = \frac{1}{2}cos²(0) = \frac{1}{2} and P_VH = P_HV = \frac{1}{2}sin²(0) = 0) and completely anti-correlated outcomes for α-β=\frac{∏}{2} (P_VV = P_HH = \frac{1}{2}cos²(\frac{∏}{2}) = 0 and P_VH = P_HV = \frac{1}{2}sin²(\frac{∏}{2}) = \frac{1}{2}).

Instead of theta = (a-b), shouldn't that be (b-a)? So E(a,b) = cos^2(b-a)?

It doesn't matter, the calculation is the same either way. Your equation for E(a,b) is wrong. E(a,b) = cos²(a-b) - sin²(a-b) = cos2(a-b). That's why your computations of E in the following are wrong.

How do you get theta = -45 from (-45, -25.5)? Shouldn't that be theta = (-25.5 - (-45)) = 25.5? Also theta = -135 from (-45, 22.5)?

This is what I get:

E(a,b) = E(-45, -25.5) = cos^2(19.5) = 0.89
E(a,b') = E(-45, 25.5) = cos^2(70.5) = 0.11
E(a',b) = E(0, -25.5) = cos^2(-25.5) = 0.82
E(a',b') = E(0, 25.5) = cos^2(25.5) = 0.82

S = 0.89 - 0.11 + 0.82 + 0.82 = 2.42

E(a,b) = cos2(a-b) = cos2(-45 + 22.5) = cos(-45) = 1/√2
Etc.

 

[DrChinese]

This is what I get:

E(a,b) = E(-45, -25.5) = cos^2(19.5) = 0.89
E(a,b') = E(-45, 25.5) = cos^2(70.5) = 0.11
E(a',b) = E(0, -25.5) = cos^2(-25.5) = 0.82
E(a',b') = E(0, 25.5) = cos^2(25.5) = 0.82

S = 0.89 - 0.11 + 0.82 + 0.82 = 2.42

You used the wrong angle settings. Instead of 25.5 it should be 22.5. Then the arithmetic works out.

Please stop referring to Malus and classical predictions for entanglement, a non-classical phenomena. This is your last warning on that.

At this point, it is completely unclear what question you are asking. You are switching from context to context so rapidly that none of the responders can assist. So ask a question clearly and immediately that can be addressed, or I suspect this thread will be reaching a conclusion. If you answer anything else instead, please expect the obvious to occur. This is not a forum for debate, or for you to put forth your (misinformed) opinions. We have been quite patient, Jabbu, but that patience is wearing thin. There are multiple professional physicists and science advisors spending time here, and you are going in circles.

 

[stevendaryl]

What you showed are some statements that do not correspond to anything I've seen anywhere else. You used no any equations, you just asserted "we can prove" without showing any proof.

The predictions of QM for a single angle \theta=0 using entangled photons is this:
(Both Alice and Bob keep their filters set at this angle, for each trial).

50\% of the time, one photon passes through Alice's filter and the other passes through Bob's filter.[/itex]
50\% of the time, one photon is blocked by Alice's filter and the other is blocked by Bob's filter.
It never happens that Alice and Bob get different results.

Do you understand that those are the predictions of QM for this case?

Do you really not understand how one could duplicate those predictions without nonlocal interactions?

Suppose that instead of photons, we have slips of paper that messages are written on. Instead of Alice's filter, we have a person who reads one of the slips of paper, and either says "Pass" or "Block". Do you really not see how you could generate slips of paper so that we could guarantee that Alice gets "Pass" 50% of the time, and Bob gets "Pass" 50% of the time, and they always get the same result?

Of course, you can do it. Half of the time, you write "Pass" on both pieces of paper. Half the time you write "Block" on both pieces of paper. That DOES it! It's trivial.

The weird fact is that if instead of one filter setting, Alice and Bob have a choice of three settings, then there is NO way to do it using messages written on pieces of paper.

 

[stevendaryl]

The predictions of QM for a single angle \theta=0 using entangled photons is this:
(Both Alice and Bob keep their filters set at this angle, for each trial).

50\% of the time, one photon passes through Alice's filter and the other passes through Bob's filter.[/itex]
50\% of the time, one photon is blocked by Alice's filter and the other is blocked by Bob's filter.
It never happens that Alice and Bob get different results.

Do you understand that those are the predictions of QM for this case?

Do you really not understand how one could duplicate those predictions without nonlocal interactions?

Suppose that instead of photons, we have slips of paper that messages are written on. Instead of Alice's filter, we have a person who reads one of the slips of paper, and either says "Pass" or "Block". Do you really not see how you could generate slips of paper so that we could guarantee that Alice gets "Pass" 50% of the time, and Bob gets "Pass" 50% of the time, and they always get the same result?

Of course, you can do it. Half of the time, you write "Pass" on both pieces of paper. Half the time you write "Block" on both pieces of paper. That DOES it! It's trivial.

The weird fact is that if instead of one filter setting, Alice and Bob have a choice of three settings, then there is NO way to do it using messages written on pieces of paper.

Jabbu, can you just say whether you agree that the thought experiment in terms of message passing on pieces of paper can be implemented to give the same statistical predictions as QM, in the case of a single filter setting? That conclusion is so trivial, that I really have no idea what you're asking for when you ask for a proof of it.

The harder proof is to show that it CAN'T be done with three filter settings.

 

[stevendaryl]

Because that's how classical physics prediction is calculated. What do you think is classical physics prediction for theta = 0 degrees?

This is not about classical physics. It's about quantum physics. Malus' law has nothing to do with the argument. Absolutely nothing. Your bringing up Malus' law over and over again means that you really, really don't understand what people are saying to you.

 

[Jabbu]

I don't know what you're talking about, sorry. You have completely correlated outcomes for α-β = 0 (P_VV = P_HH = \frac{1}{2}cos²(0) = \frac{1}{2} and P_VH = P_HV = \frac{1}{2}sin²(0) = 0) and completely anti-correlated outcomes for α-β=\frac{∏}{2} (P_VV = P_HH = \frac{1}{2}cos²(\frac{∏}{2}) = 0 and P_VH = P_HV = \frac{1}{2}sin²(\frac{∏}{2}) = \frac{1}{2}).

I'm talking about those terms in equation 20, in an experiment they are not calculate but counted.

It doesn't matter, the calculation is the same either way. Your equation for E(a,b) is wrong. E(a,b) = cos²(a-b) - sin²(a-b) = cos2(a-b). That's why your computations of E in the following are wrong.

If by "cos2(x)" you don't mean to say "cos^2(x)" I think 2 should go inside brackets like this "cos(2x)".

E(a,b) = cos2(a-b) = cos2(-45 + 22.5) = cos(-45) = 1/√2
Etc.

Ok. So how cos^2(theta) fits in? When a= -30 and b= +30 QM predicts correlation = cos^2(60) = 25%, right? So isn't E(-30,30) supposed to be that same "correlation" value?

 

[Jabbu]

You used the wrong angle settings. Instead of 25.5 it should be 22.5.

Thanks.

Please stop referring to Malus and classical predictions for entanglement, a non-classical phenomena.

How else do you compare experimental results with classical physics prediction?

At this point, it is completely unclear what question you are asking.

I'm talking to RUTA. See the paper he posted, or please ask specific questions about whatever it is unclear to you.

 

[Jabbu]

This is not about classical physics. It's about quantum physics. Malus' law has nothing to do with the argument. Absolutely nothing. Your bringing up Malus' law over and over again means that you really, really don't understand what people are saying to you.

To understand practical implications it is paramount to understand how experimental results differ from classical prediction. Please read the paper RUTA posted, or any other paper or article about Bell's theorem. Comparing experimental results with classical prediction is very important part of the analysis.

 

[Jabbu]

Jabbu, can you just say whether you agree that the thought experiment in terms of message passing on pieces of paper can be implemented to give the same statistical predictions as QM, in the case of a single filter setting?

To be more specific instead of hypothetical piece of paper it's better to attribute hidden variables as properties of actual entities, in this case photons and polarizers. Then it's easier to realize real conditions these variables have to satisfy. So to answer your question, unless the paper says "cos^2(theta)" it would be refuted by every other experiment with known relative polarization different than 45 degrees.

But if you insist your hidden variable can fake experimental results for a single theta setting, then you need to realize S = E(a,b) - E(a,b') + E(a',b) + E(a',b') is defined by independent results of four separate theta settings from four separate experiments. Therefore, if your hidden variable can fake each of those experiments individually it will automatically fake the value of S.

http://en.wikipedia.org/wiki/CHSH_inequality

 

[atyy]

To be more specific instead of hypothetical piece of paper it's better to attribute hidden variables as properties of actual entities, in this case photons and polarizers. Then it's easier to realize real conditions these variables have to satisfy. So to answer your question, unless the paper says "cos^2(theta)" it would be refuted by every other experiment with known relative polarization different than 45 degrees.

Bell's theorem can also rule out any modification of the classical laws that are local, so it doesn't just rule out Malus's Law acting on classical photon pairs, but all local alternatives.

Nonetheless, if you would like to try to see if how close Malus's Law and classical photon pairs can come to mimicking the quantum entangled pairs, you can try this. First consider that each side receives 50% vertical and 50% horizontal photons, and that when their polarizers are both vertical, both sides always get the same result. In this case, you can imagine that this result is obtained because the classical source sends out 50% classical pairs with both photons vertically polarized, and 50% classical pairs with both photons horizontally polarized.

Then keeping the same assumption about the classical source, you can apply Malus's Law for other polarizer settings, and carry out the analysis for this classical case to compare with the quantum entangled case.

 

[stevendaryl]

To be more specific instead of hypothetical piece of paper it's better to attribute hidden variables as properties of actual entities, in this case photons and polarizers.

But that would be a silly thing to do. We know that the classical theory of polarization cannot explain the results of the EPR experiment for entangled photons. We already know that. Malus' law does not describe the situation. That's completely clear.

The issue is whether some other law describes the EPR experiment in terms of local interactions. That's the question that Bell was interested int.

I don't know why you keep bringing up Malus' law. We know that Malus' law doesn't work in the case of entangled photons.

Then it's easier to realize real conditions these variables have to satisfy. So to answer your question, unless the paper says "cos^2(theta)" it would be refuted by every other experiment with known relative polarization different than 45 degrees.

In the case of entangled photons, there is no polarization angle of the photons. It's completely unknown. QM actually says that the photons don't have a polarization.

As I said, you keep bringing up Malus' law when it has nothing to do with the argument that is being made. If you don't know the polarization angles, then you can't apply Malus' law.

What you can do is to assume that there is some variable associated with the pair of photons that determines whether it is absorbed or transmitted by a polarizing filter. In the case of a single polarization angle, that is possible. In the case of 3 different possible angles, it is not possible.

You ask a question: Why 3 angles. Then you don't pay any attention to the answer. It's very frustrating.

 

[DrChinese]

1. How else do you compare experimental results with classical physics prediction?

2. I'm talking to RUTA. See the paper he posted, or please ask specific questions about whatever it is unclear to you.

1. There are no classical predictions for entanglement. There is no such thing as entanglement in the classical world. EPR (1935). We have discussed this in the other threads that are now closed. In this thread, I showed you in great detail how any classical attempt to achieve the entangled state statistics would fail. Rather than stop there, you have kept on and wasted the time of others.

2. No, you are arguing with RUTA, me, atyy, Nugatory, bhobba, stevendaryl and probably a few others. I am sorry, but you are going around in circles again and clearly have no question out on the table to get answered.

 

[Jabbu]

Bell's theorem can also rule out any modification of the classical laws that are local, so it doesn't just rule out Malus's Law acting on classical photon pairs, but all local alternatives.

Yes, and look at it from the other side. As much as the inequality sets the boundaries from one side, so do classical physics laws set their own limits on the other side, as usual. The hidden variable is in a very tight spot, it has to sneak in between the both, and it doesn't really make sense to suggest local theories if they describe reality more bizarre than non-locality itself.

Nonetheless, if you would like to try to see if how close Malus's Law and classical photon pairs can come to mimicking the quantum entangled pairs, you can try this. First consider that each side receives 50% vertical and 50% horizontal photons, and that when their polarizers are both vertical, both sides always get the same result. In this case, you can imagine that this result is obtained because the classical source sends out 50% classical pairs with both photons are vertically polarized, and 50% classical pairs with both photons horizontally polarized.

The paper posted by RUTA says this:

- "In each pair, the signal and idler photon have the same polarization Ls = Li = L. As successive pairs are produced, L changes in an unpredictable manner that uniformly covers the whole range of possible polarizations. The quantity L is the ‘‘hidden variable,’’ a piece of information that is absent from quantum mechanics."

It seems L (lambda) directly corresponds only to unknown photon polarization, and since they say it uniformly covers the whole range, that should be the same as unpolarized light. Malus can make predictions only if photon polarization relative to the polarizer is known, or known to be uniformly random, as the paper suggest, in which case the probability is 50-50% as you said. That's 0% correlation for any Alice and Bob polarizer settings. Classically then, there is simply no reason for two separate uniformly random events to be correlated, at all.

 

[stevendaryl]

But if you insist your hidden variable can fake experimental results for a single theta setting, then you need to realize S = E(a,b) - E(a,b') + E(a',b) + E(a',b') is defined by independent results of four separate theta settings from four separate experiments. Therefore, if your hidden variable can fake each of those experiments individually it will automatically fake the value of S.

No, that's not true. I'm getting very frustrated with you, because you keep asking questions, and don't seem interested in the answers.

Let me try one more time to see if I can explain to you why 4 angles (or 3) makes a difference.

We can come up with a hidden-variables model that reproduces the QM predictions for angles a and b. We can come up with a DIFFERENT model that reproduces the predictions for angles a and b'. We can come up with a third model that reproduces the predictions for angles a' and b. We can come up with a fourth model that reproduces the predictions for a' and b'.

What we can't do is to come up with a SINGLE[/itex] model that works in all 4 cases. That's the point! There is no way to combine the 4 models into one.

The way that a hidden-variables model of EPR would work is the following:

1. When the pair of photons is created, there is some hidden information \lambda associated with each photon. This extra information is NOT polarization angle, because it's easy to see that Malus' law is not sufficient to explain EPR.
2. When Alice's photon reaches her filter, her filter is at some angle \theta_A.
3. We assume that whether it passes or not is a function: f_A(\lambda, \theta_A) that depends on both the extra information in the photon, \lambda, and on the angle of Alice's filter, \theta_A. The function f_A returns either +1 (meaning the photon passes) or -1 (meaning the photon is blocked).
4. Similarly, when Bob's photon reaches his filter, whether it passes or not is dependent on a different function f_B(\lambda, \theta_B)

The point is that the hidden variable \lambda has to be chosen BEFORE[/itex] it is known what angles Alice and Bob will choose. That makes it a lot harder. If Alice and Bob's angles are known ahead of time, then it's easy to come up with a \lambda that works.

That's why multiple angles is harder: You have to choose \lambda that works with any possible choice made by Alice and Bob.

 

[Jabbu]

...and clearly have no question out on the table to get answered.

We are not going in circles, we came all the way to talk about these two equations:

E(a,b) = Pvv(a,b) + Phh(a,b) - Pvh(a,b) - Phv(a,b)

S = E(a,b) - E(a,b') + E(a',b) + E(a',b')


My questions are clearly marked with question marks: - So how cos^2(theta) fits in? When a= -30 and b= +30 QM predicts correlation = cos^2(60) = 25%, right? So isn't E(-30,30) supposed to be that same "correlation" value?

 

[stevendaryl]

We are not going in circles, we came all the way to talk about these two equations:

E(a,b) = Pvv(a,b) + Phh(a,b) - Pvh(a,b) - Phv(a,b)

S = E(a,b) - E(a,b') + E(a',b) + E(a',b')


My questions are clearly marked with question marks: - So how cos^2(theta) fits in? When a= -30 and b= +30 QM predicts correlation = cos^2(60) = 25%, right? So isn't E(-30,30) supposed to be that same "correlation" value?

No, E(a,b) is defined to be (you've got the definition right there):

E(a,b) = P_{vv}(a,b) +P_{hh}(a,b) - P_{hv}(a,b) - P_{hv}(a,b)

The relationship with cos^2(\theta) is this:
P_{vv}(a,b) +P_{hh}(a,b) = cos^2(a - b)
P_{vh}(a,b) +P_{hv}(a,b) = 1 - cos^2(a - b) = sin^2(a-b)

So E(a,b) = cos^2(a-b) - sin^2(a-b)

 

[atyy]

Yes, and look at it from the other side. As much as the inequality sets the boundaries from one side, so do classical physics laws set their own limits on the other side, as usual. The hidden variable is in a very tight spot, it has to sneak in between the both, and it doesn't really make sense to suggest local theories if they describe reality more bizarre than non-locality itself.

"Bizarre" is a matter of taste. Also, maybe one day quantum mechanics will be falsified, and we will have to look for a new theory. Bell's theorem says that even if one finds quantum mechanics too bizarre, or if quantum mechanics is experimentally falsified, the theory that replaces quantum mechanics must also be nonlocal. (Caveat: there remain a couple of loopholes in the experimental implementations of the Bell tests.)

The paper posted by RUTA says this:

- "In each pair, the signal and idler photon have the same polarization Ls = Li = L. As successive pairs are produced, L changes in an unpredictable manner that uniformly covers the whole range of possible polarizations. The quantity L is the ‘‘hidden variable,’’ a piece of information that is absent from quantum mechanics."

It seems L (lambda) directly corresponds only to unknown photon polarization, and since they say it uniformly covers the whole range, that should be the same as unpolarized light. Malus can make predictions only if photon polarization relative to the polarizer is known, or known to be uniformly random, as the paper suggest, in which case the probability is 50-50% as you said. That's 0% correlation for any Alice and Bob polarizer settings. Classically then, there is simply no reason for two separate uniformly random events to be correlated, at all.

In that paper http://arxiv.org/abs/quant-ph/0205171, the local hidden variable theory they propose does not assume Malus's Law, instead they replace it with their own law (Eq 18), because they want to propose a local hidden variable theory that achieves the limit allowed by one of the Bell inequalities (Eq 22). If we used Malus's Law (instead of Eq 18), then we would have P_V(γ,λ) = cos²(γ-λ). When I put Malus's Law into their Eq 19 for both polarizers vertically oriented, I get P_VV(0,0) = 3/8. If there were no correlation between Alice and Bob, then I would expect P_VV(0,0) = 1/4. I'm not sure I did that right, but it seems that with their local hidden variable theory does produce more than 0% correlation between Alice and Bob. This correlation is due to a local hidden variable, which is built in at the source - although the polarization of the photons is random, both photons in each pair have the same correlation.

 

[RUTA]

I'm talking about those terms in equation 20, in an experiment they are not calculate but counted.

Yes, there are coincidence counts for data in the next part of the paper, but I didn't see that what you wrote had anything to do with experimental data of that type.

If by "cos2(x)" you don't mean to say "cos^2(x)" I think 2 should go inside brackets like this "cos(2x)".

I meant cos(2x) per the trig identity I showed you.

Ok. So how cos^2(theta) fits in? When a= -30 and b= +30 QM predicts correlation = cos^2(60) = 25%, right? So isn't E(-30,30) supposed to be that same "correlation" value?

No, you're confusing P_VV with E.

 

[johana]

What we can't do is to come up with a SINGLE[/itex] model that works in all 4 cases. That's the point! There is no way to combine the 4 models into one.

That doesn't answer why would they test 4 cases in the same experiment, instead of individually in four separate experiments.

The point is that the hidden variable \lambda has to be chosen BEFORE[/itex] it is known what angles Alice and Bob will choose. That makes it a lot harder. If Alice and Bob's angles are known ahead of time, then it's easy to come up with a \lambda that works.

Let so called "Bell test angles" be known ahead of time: 0°, 45°, 22.5° and 67.5°. How that makes it any easier to come up with a single \lambda function that works for each combination: E(a,b), E(a,b'), E(a',b), and E(a',b')? What time has to do with any of it?

 

[Nugatory]

That doesn't answer why would they test 4 cases in the same experiment, instead of individually in four separate experiments.

We're comparing coincidence rates at various angles. Thus we need a setup that gives us a series of measurements in which everything is the same except the angles. The most practical way of doing that is to run one experiment in which only the angle varies.

Let so called "Bell test angles" be known ahead of time: 0°, 45°, 22.5° and 67.5°. How that makes it any easier to come up with a single \lambda function that works for each combination: E(a,b), E(a,b'), E(a',b), and E(a',b')?

it doesn't, and that's why we use multiple angles. It's easy to come up with a $\lambda$ function that matches the quantum mechanical prediction if we know the two angles that any given photon pair will encounter, but impossible if we only know up front that the pair will encounter some combination of two of those four angles.

 

[stevendaryl]

Let so called "Bell test angles" be known ahead of time: 0°, 45°, 22.5° and 67.5°. How that makes it any easier to come up with a single \lambda function that works for each combination: E(a,b), E(a,b'), E(a',b), and E(a',b')? What time has to do with any of it?

If you know the angles ahead of time, it is easy to reproduce the predicted correlations using a local hidden-variables model.

If you know ahead of time that Alice's filter is set at angle a and Bob's filter is set at angle b, then a model that reproduces the predictions of QM is the following:

1. With probability \frac{1}{2} cos^2(a-b), send a photon to Alice that is polarized at angle a, and send a photon to Bob that is polarized at angle b.
2. With probability \frac{1}{2} cos^2(a-b), send a photon to Alice that is polarized at angle a+90^o, and send a photon to Bob that is polarized at angle b+90^o.
3. With probability \frac{1}{2} sin^2(a-b), send a photon to Alice that is polarized at angle a, and send a photon to Bob that is polarized at angle b+90^o.
4. With probability \frac{1}{2} sin^2(a-b), send a photon to Alice that is polarized at angle a+90^o, and send a photon to Bob that is polarized at angle b.

We can independently verify that if a filter is aligned in the same direction as polarized light, then it passes 100% of the time, and if it is aligned at a 90 degree angle, relative to the polarized light, then it is blocked 100% of the time.

This trivial model reproduces exactly the predictions of QM for the twin-photon EPR experiment.

It's clear that the model could not work if you don't know Alice's and Bob's filter orientations ahead of time.

 

[johana]

We're comparing coincidence rates at various angles. Thus we need a setup that gives us a series of measurements in which everything is the same except the angles. The most practical way of doing that is to run one experiment in which only the angle varies.

What is the point of randomly switching angles and taking measurements "simultaneously" little by little, instead of to test each angle completely and separately one after another?

 

[atyy]

What is the point of randomly switching angles and taking measurements "simultaneously" little by little, instead of to test each angle completely and separately one after another?

In the Bell inequalities, it is assumed (1) that Bob's result only depends on his choice and the local hidden variables, and it is also assumed (2) that Bob's result cannot depend on Alice's choice. The idea behind making the choices randomly at each side is that the choice is only made at the "last moment" before the the result is obtained, so that if we assume that the speed of light is an upper limit to the speed of communication, Alice's choice cannot be communicated in time to Bob to affect his result.

 

[DrChinese]

What is the point of randomly switching angles and taking measurements "simultaneously" little by little, instead of to test each angle completely and separately one after another?

There is nothing wrong with that. Most Bell tests do just that.

The purpose of those that do fast switching is to demonstrate that there is no additional sub-lightspeed action in play that might affect the results. But we already know that from experiments such as Weihs et al, so that is no longer relevant. atyy's post above covers this issue.

 

[johana]

If you know the angles ahead of time, it is easy to reproduce the predicted correlations using a local hidden-variables model.

If you know ahead of time that Alice's filter is set at angle a and Bob's filter is set at angle b, then a model that reproduces the predictions of QM is the following:

1. With probability \frac{1}{2} cos^2(a-b), send a photon to Alice that is polarized at angle a, and send a photon to Bob that is polarized at angle b.
2. With probability \frac{1}{2} cos^2(a-b), send a photon to Alice that is polarized at angle a+90^o, and send a photon to Bob that is polarized at angle b+90^o.
3. With probability \frac{1}{2} sin^2(a-b), send a photon to Alice that is polarized at angle a, and send a photon to Bob that is polarized at angle b+90^o.
4. With probability \frac{1}{2} sin^2(a-b), send a photon to Alice that is polarized at angle a+90^o, and send a photon to Bob that is polarized at angle b.

Those are four different functions. As you said the whole point is to come up with a single model, can you do that?

You can not use (a-b) for a local hidden variable, that's exactly what makes QM equations nonlocal. In local reality properties and interaction of photon B and polarizer B are of no consequence to photon A and polarizer A, and vice versa. For local reality a hidden variable must fit a single function that applies separately to each side.

For each photon pair:

\lambda(photon_A, polarizer_A) = V|H
\lambda(photon_B, polarizer_B) = V|H

...which integrated over many pairs sums up to yield:

VV + HH - VH - HV = cos^2(A-B) - sin^2(A-B)

 

[stevendaryl]

Those are four different functions. As you said the whole point is to come up with a single model, can you do that?

It's not 4 different functions, it's 4 different values for the hidden variable \lambda. If \lambda = \lambda_1, both Alice's and Bob's photons pass their filters. This value is chosen with probability \frac{1}{2} cos^2(a-b), etc.

You can not use (a-b) for a local hidden variable, that's exactly what makes QM equations nonlocal.

That was the point I was making. If a and b are known ahead of time, then there is nothing nonlocal involved in taking those values into account. It only becomes nonlocal if you allow Alice and Bob to choose a and b while the photons are in flight.

 

[stevendaryl]

That was the point I was making. If a and b are known ahead of time, then there is nothing nonlocal involved in taking those values into account. It only becomes nonlocal if you allow Alice and Bob to choose a and b while the photons are in flight.

I should say: If the values of a and b are known ahead of time, it's not necessarily nonlocal.

 

[johana]

In the Bell inequalities, it is assumed (1) that Bob's result only depends on his choice and the local hidden variables, and it is also assumed (2) that Bob's result cannot depend on Alice's choice. The idea behind making the choices randomly at each side is that the choice is only made at the "last moment" before the the result is obtained, so that if we assume that the speed of light is an upper limit to the speed of communication, Alice's choice cannot be communicated in time to Bob to affect his result.

FTL question is addressed by placing the two polarizers far enough apart so the time difference between when "signal" and "idler" photon meet with their polarizers is less than the speed of light would require to go from one to the other.

Let signal photon A go vertical through polarizer A at time t= 0. Let idler photon B be one light year away from photon A and one meter away from polarizer B at time t= 0. Photon B now has to assume the same polarization as photon A before it meets with polarizer B, that's the trick regarding FTL. But what angle polarizer B is set to, and when, is of no consequence to this speed of correlation/information between photon A and photon B.

 

[johana]

It's not 4 different functions, it's 4 different values for the hidden variable \lambda. If \lambda = \lambda_1, both Alice's and Bob's photons pass their filters. This value is chosen with probability \frac{1}{2} cos^2(a-b), etc.

Proposing different models for each angle is invalid to begin with. Plus, you can not use (a-b) to define a hidden local variable. It does not qualify, it does not compare.

That was the point I was making. If a and b are known ahead of time, then there is nothing nonlocal involved in taking those values into account. It only becomes nonlocal if you allow Alice and Bob to choose a and b while the photons are in flight.

Angles are always known and deliberately chosen ahead of time. Are you saying there is some difference if we randomly switch angles and take measurements "simultaneously" little by little, instead of to test each angle completely and separately one after another?

 

[atyy]

FTL question is addressed by placing the two polarizers far enough apart so the time difference between when "signal" and "idler" photon meet with their polarizers is less than the speed of light would require to go from one to the other.

Let signal photon A go vertical through polarizer A at time t= 0. Let idler photon B be one light year away from photon A and one meter away from polarizer B at time t= 0. Photon B now has to assume the same polarization as photon A before it meets with polarizer B, that's the trick regarding FTL. But what angle polarizer B is set to, and when, is of no consequence to this speed of correlation/information between photon A and photon B.

Take a look at stevendaryl's post #71. If the settings are known ahead of time, then his variable (a-b) can be a local hidden variable, ie. although (a-b) is associated with distant apparatuses, there's plenty of time to propagate the detector settings back to the source.

 

[stevendaryl]

Proposing different models for each angle is invalid to begin with. Plus, you can not use (a-b) to define a hidden local variable. It does not qualify, it does not compare.

Sigh. You're coming into the middle of a long discussion, and your points are not relevant to the particular point I was discussing. If the order of the events is the following:
(I'm saying IF it takes place in the following way---I'm not saying that it does, and I'm not
saying that Bell allowed for it, or whatever. This is simply a hypothetical explanation for a particular thought experiment.)

1. Alice chooses her filter angle a
2. Bob chooses his filter angle b
3. A hidden variable \lambda is generated taking into account a and b.
4. A pair of photons is generated that somehow encode this value
5. Whether Alice's photon passes or not is a deterministic function of \lambda and a
6. Whether Bob's photon passes or not is a deterministic function of \lambda and b.

My claim is that it is possible to implement the above scenario, using local hidden variables, in a way that reproduces the predictions of quantum mechanics. If the choice of \lambda depends on the settings a and b, then there is no proof of nonlocality (or nonrealism, or whatever it is that Bell's theorem talks about).

To get Bell's proof to go through, you have to assume that the hidden variable \lambda is NOT dependent on Alice's and Bob's filter settings. If you don't make that assumption, then there is certainly a local hidden-variables explanation.

Angles are always known and deliberately chosen ahead of time. Are you saying there is some difference if we randomly switch angles and take measurements "simultaneously" little by little, instead of to test each angle completely and separately one after another?

If the settings are chosen ahead of time, then that leaves a loophole for local hidden variables theories. It might not be a plausible loophole, but it's a loophole.

 

[DrChinese]

johana,

Just to make sure everyone is on the same page: no one is arguing that it makes a difference if there is fast switching or not. However, that is something we know now that was not known 100% prior to experiment (Aspect, Weihs, etc.). If you want to rule out a local action so that the settings A and B are provably independent, use fast switching. So stevendaryl is simply saying that a local realistic model was technically viable prior to that because there might be time for subluminal communication from one measuring device to the other (crazy as it seems). But now we know better.

 

[stevendaryl]

johana,

Just to make sure everyone is on the same page: no one is arguing that it makes a difference if there is fast switching or not. However, that is something we know now that was not known 100% prior to experiment (Aspect, Weihs, etc.). If you want to rule out a local action so that the settings A and B are provably independent, use fast switching. So stevendaryl is simply saying that a local realistic model was technically viable prior to that because there might be time for subluminal communication from one measuring device to the other (crazy as it seems). But now we know better.

Exactly. Thanks.

 

[johana]

If the settings are chosen ahead of time, then that leaves a loophole for local hidden variables theories. It might not be a plausible loophole, but it's a loophole.

Yes, loophole. But what loophole are you talking about? FTL loophole is addressed by separating polarizers far enough apart relative to the time interval between when signal and idler photon are supposed to meet with their polarizers. It works for any individual angle when tested separately, so that's not it.

 

[johana]

Take a look at stevendaryl's post #71. If the settings are known ahead of time, then his variable (a-b) can be a local hidden variable, ie. although (a-b) is associated with distant apparatuses, there's plenty of time to propagate the detector settings back to the source.

time0: photons A & B emitted with unknown/undefined polarization

time1: photon A goes through 0° polarizer A and acquires 0° polarization

time2: photon B acquires 0° polarization, for some reason ( time1 = time2 ?? )

time3: photon B with 0° polarization interacts with 0° polarizer B, so it too goes throughNow, if polarizer B was at 90° and switched to 0° just a moment before time3, what difference does it make?

 

[Nugatory]

Now, if polarizer B was at 90° just a moment before time3, what difference does it make?

Not much. But suppose that polarizer A was at 90° until just a moment before $t1$... In fact, for such a short moment that both $t2$ and $t3$ have come and gone before a light signal from polarizer A's flip to 0 degrees could have reached photon B... That makes B's flip to zero much harder to explain.

 

[atyy]

Now, if polarizer B was at 90° and switched to 0° just a moment before time3, what difference does it make?

To add to Nugatory's point, you can read more discussion of the issue in:
http://physics.aps.org/synopsis-for/10.1103/PhysRevLett.105.250404
http://arxiv.org/abs/1007.5518
http://arxiv.org/abs/1008.3612
http://arxiv.org/abs/1303.2849 (section B.2 "Locality loophole")

 

[johana]

Not much. But suppose that polarizer A was at 90° until just a moment before $t1$... In fact, for such a short moment that both $t2$ and $t3$ have come and gone before a light signal from polarizer A's flip to 0 degrees could have reached photon B... That makes B's flip to zero much harder to explain.

It is accepted by both sides time2 is less than time3, otherwise causality would be acting backwards in time, and that's not really kind of thing local realists are hoping for. Each subsequent time is greater or equal to previous time:

time0: photons A & B emitted with unknown/undefined polarization

->time0.7: polarizer A set to 45°
->time0.8: polarizer B set to 90°

time0.9: polarizer A set to 0°
time1: photon A goes through 0° polarizer A and acquires 0° polarization

->time1.7: polarizer A set to 15°
->time1.8: polarizer B set to 75°

time2: photon B acquires 0° polarization, for some reason ( time1 = time2 ?? )

->time2.7: polarizer A set to 90°
->time2.8: polarizer B set to 45°

time2.9: polarizer B set to 0°
time3: photon B with 0° polarization interacts with 0° polarizer B, so it too goes through


Please note those time events marked with arrows, are they anyhow relevant to what happens at time1, time2, or time3?

 

[StevieTNZ]

Isn't it when the photon is detected by an apparatus, it is then that the system takes on the polarisation of the polariser?

 

[johana]

To add to Nugatory's point, you can read more discussion of the issue in:
http://physics.aps.org/synopsis-for/10.1103/PhysRevLett.105.250404
http://arxiv.org/abs/1007.5518
http://arxiv.org/abs/1008.3612
http://arxiv.org/abs/1303.2849 (section B.2 "Locality loophole")

They do talk about it, but they don't explain it. Try to answer the question in your own words, with a sentence or two, and maybe you'll see the same paradox I see. Or maybe you'll see an explanation, that's even better.

 

[atyy]

They do talk about it, but they don't explain it. Try to answer the question in your own words, with a sentence or two, and maybe you'll see the same paradox I see. Or maybe you'll see an explanation, that's even better.

They do. Read the papers.

 

[johana]

Isn't it when the photon is detected by an apparatus, it is then that the system takes on the polarisation of the polariser?

It is when the first photon A interacts with the polarizer A that its "wave function collapses", which we are meant to understand its, until then, "undefined" polarization becomes real and defined by the polarizer A. At the same time, or a bit later, the other photon B, whose polarization was also undefined until then, acquires the same definite polarization as the first photon A. From there on both photons are in "classic" mode of operation, having definite polarization, and so the interaction between second photon B and polarizer B goes on normally in accordance to Malus law.

 

[stevendaryl]

Yes, loophole. But what loophole are you talking about? FTL loophole is addressed by separating polarizers far enough apart relative to the time interval between when signal and idler photon are supposed to meet with their polarizers. It works for any individual angle when tested separately, so that's not it.

No, that is it. If angles a and b are chosen too far in advance, then you can't be certain that the hidden variable isn't influenced by those settings.

 

[stevendaryl]

They do talk about it, but they don't explain it. Try to answer the question in your own words, with a sentence or two, and maybe you'll see the same paradox I see. Or maybe you'll see an explanation, that's even better.

No, I don't know what paradox you are talking about.

 

[stevendaryl]

It is when the first photon A interacts with the polarizer A that its "wave function collapses", which we are meant to understand its, until then, "undefined" polarization becomes real and defined by the polarizer A. At the same time, or a bit later, the other photon B, whose polarization was also undefined until then, acquires the same definite polarization as the first photon A. From there on both photons are in "classic" mode of operation, having definite polarization, and so the interaction between second photon B and polarizer B goes on normally in accordance to Malus law.

That's the "collapse" interpretation of EPR, but it's explicitly nonlocal.

 

[Nugatory]

It is accepted by both sides time2 is less than time3, otherwise causality would be acting backwards in time, and that's not really kind of thing local realists are hoping for. Each subsequent time is greater or equal to previous time:

Time1 is not necessarily less than time3. You can set up the experiment in such a way that some of the people watching the experiment observe that photon A reaches polarizer A before photon B reaches polarizer B while others (who happen to be moving relative to the first group) observe that photon B reaches polarizer B before photon A reaches polarizer A. This is Einstein's "relativity of simultaneity" (google for that phrase if you're not already familiar with it) at work.

Because we cannot directly observe photon B acquiring 0° polarization, we don't know if time2 is different from time3 (it's natural to expect that it is, but that expectation comes from our day-to-day experience, which is a poor guide to how QM works). However, if time2 is different from time3, it will always be less than time3 for all observers.

Please note those time events marked with arrows, are they anyhow relevant to what happens at time1, time2, or time3?

As long as they are all greater than time0 the changes to the A setting cannot affect the result at B (and vice versa) in any local theory. Yet the quantum mechanical result is that the result at B depends on the setting at A at time0.9 - the reason for doing the fast switching experiments is to confirm that this is indeed the case.

 

[RUTA]

time0: photons A & B emitted with unknown/undefined polarization

time1: photon A goes through 0° polarizer A and acquires 0° polarization

time2: photon B acquires 0° polarization, for some reason ( time1 = time2 ?? )

time3: photon B with 0° polarization interacts with 0° polarizer B, so it too goes through


Now, if polarizer B was at 90° and switched to 0° just a moment before time3, what difference does it make?

Let's look at the Hardy state from the Mermin paper in my post #21. You have two particles, two settings 1 and 2, two outcomes R and G. Here are the facts for that state being measured in this configuration taken from p 881:

The data exhibit the following important features:

(a) In runs in which the detectors end up with different
settings, they never both flash green: 21 GG and
12 GG never occur.

(b) In runs in which both detectors end up set to 2, one
occasionally finds both flashing green: 22GG sometimes
occurs.

(c) In runs in which both detectors end up set to 1, they
never both flash red: 11RR never occurs.

Now suppose you're particle 1 and know ahead of time that you both will be measured in setting 2. You agree with your partner that you'll both give a G result, i.e., 22GG. As you approach your detector you see that it's set to 1. What do you do?

Suppose you choose to be 1R, thinking your partner will be 2G and you have to satisfy condition (a). But, what happens if your partner also encounters setting 1 and, thinking the same thing, decides to be 1R. Now you have a 11RR outcome in violation of (c). Suppose you go with 1G. Now if your partner's detector isn't changed and he thinks all is ok and goes with 2G you have a 12GG outcome in violation of (a). So, by encountering a setting that isn't what you expected you're screwed.

Is that clear enough?

 

[stevendaryl]

Let's look at the Hardy state from the Mermin paper in my post #21. You have two particles, two settings 1 and 2, two outcomes R and G.

Yes. In my opinion, even though Bell's inequality might be more amenable to experimental tests, when it comes to discussions about the weirdness of quantum mechanics and entanglement, the Hardy state is much easier to understand. The impossibility of a local, deterministic hidden-variables explanation is much clearer, since you don't really need to do any kind of mathematics involving expectation values, or probability calculations, or whatever. You have a situation which clearly is impossible, classically, yet happens, quantum mechanically.

 

[Johan0001]

Suppose you choose to be 1R, thinking your partner will be 2G and you have to satisfy condition (a). But, what happens if your partner also encounters setting 1 and, thinking the same thing, decides to be 1R. Now you have a 11RR outcome in violation of (c). Suppose you go with 1G. Now if your partner's detector isn't changed and he thinks all is ok and goes with 2G you have a 12GG outcome in violation of (a). So, by encountering a setting that isn't what you expected you're screwed.

Is that clear enough?

This is the big mystery.
Could you give a reference to which experiment best exploits this phenomena, with the least possible loopholes such as reality and locality?

 

[DrChinese]

This is the big mystery.
Could you give a reference to which experiment best exploits this phenomena, with the least possible loopholes such as reality and locality?

Some of the most important include these:

http://arxiv.org/abs/quant-ph/9810080
Violation of Bell's inequality under strict Einstein locality conditions
Gregor Weihs, Thomas Jennewein, Christoph Simon, Harald Weinfurter, Anton Zeilinger (University of Innsbruck, Austria)
(Submitted on 26 Oct 1998)

We observe strong violation of Bell's inequality in an Einstein, Podolsky and Rosen type experiment with independent observers. Our experiment definitely implements the ideas behind the well known work by Aspect et al. We for the first time fully enforce the condition of locality, a central assumption in the derivation of Bell's theorem. The necessary space-like separation of the observations is achieved by sufficient physical distance between the measurement stations, by ultra-fast and random setting of the analyzers, and by completely independent data registration.

http://www.nature.com/nature/journal/v409/n6822/full/409791a0.html
Experimental violation of a Bell's inequality with efficient detection

M. A. Rowe1, D. Kielpinski1, V. Meyer1, C. A. Sackett1, W. M. Itano1, C. Monroe2 & D. J. Wineland1


http://arxiv.org/abs/1306.5772
Detection-Loophole-Free Test of Quantum Nonlocality, and Applications
B. G. Christensen, K. T. McCusker, J. Altepeter, B. Calkins, T. Gerrits, A. Lita, A. Miller, L. K. Shalm, Y. Zhang, S. W. Nam, N. Brunner, C. C. W. Lim, N. Gisin, P. G. Kwiat
(Submitted on 24 Jun 2013 (v1), last revised 26 Sep 2013 (this version, v2))

We present a source of entangled photons that violates a Bell inequality free of the "fair-sampling" assumption, by over 7 standard deviations. This violation is the first experiment with photons to close the detection loophole, and we demonstrate enough "efficiency" overhead to eventually perform a fully loophole-free test of local realism. The entanglement quality is verified by maximally violating additional Bell tests, testing the upper limit of quantum correlations. Finally, we use the source to generate secure private quantum random numbers at rates over 4 orders of magnitude beyond previous experiments.